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I was asked this question in an interview: There are two people Mike and Cheng with one blank card each. Both write either a '0' or a '1' on it without showing the other while writing. They then simultaneously show their cards to each other. If the payoffs are as follows depending on the cards shown:

1,1: Mike pays Cheng 3

0,0: Mike pays Cheng 1

1,0: Mike receives 2 from Cheng

0,1: Mike receives 2 from Cheng

Is this game fair?

Using game theory I reason: Cheng will always play 1 because his payoff from 1 is (-2 or 3) as opposed to his payoff from playing 0 which is (1 or -2). Knowing that Cheng will prefer to play 1, Mike always plays 0 to get a payoff of 2. So the game is unfair to Cheng.

But I also reason: Mike will never play 1 because he has a payoff (2 or -3) vs playing 0 which has a payoff (-1 or 2). So Mike will always prefer to play 0 (this is also inline with the above reasoning). Knowing that Mike will always prefer 0, Cheng always plays 0 so Mike gets -1 every round. So this game is unfair to Mike.

Regardless if the game is played one round or many rounds, I am unable to conclude which is correct.

Can anyone help explain why I am getting conflicting results?

PS: Later the interviewer told me that the game is unfair and Mike always wins.

PPS: I apologize in advance if I have made any mistakes in the posting - this is my first post.

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    $\begingroup$ The game you described seems to fit the definition of a zero-sum game. As such, there is not one strategy that is always best for either player (i.e. no dominant strategy, if you prefer the jargon). So your reasoning is false. $\endgroup$
    – Herr K.
    Commented Jul 6, 2017 at 7:33
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    $\begingroup$ Also, I'm not sure what definition of "fair game" your interviewer used, but I'm quite certain that he/she was wrong about "Mike always wins", especially when the two players move simultaneously. $\endgroup$
    – Herr K.
    Commented Jul 6, 2017 at 7:37
  • $\begingroup$ I think what he meant as a "fair game" was if this bet is favorable to either Mike or Cheng or neither. Having a winning strategy, I believe each player has one: Mike is always better of playing '0' because he can lose 3 if he plays '1' which is worse than losing 1 if he plays '0'. Similarly, Cheng also has a winning strategy that he is better off playing '1' because he has a chance to gain 3 vs. playing '0' and having a chance of 1. $\endgroup$
    – user13841
    Commented Jul 6, 2017 at 16:08
  • $\begingroup$ The difference is Cheng's loss is capped at -2 if he loses, whereas Mike's loss can be -3,-1. Alternatively, Mike's gain is capped at +2 if he wins, whereas Cheng's gain can be either +3 or +1. I understand dominant strategies a little but thank you for minimizing the jargon and hence have replied using 'winning' instead of 'dominant'. $\endgroup$
    – user13841
    Commented Jul 6, 2017 at 16:08
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    $\begingroup$ If you consider the following payoff matrix of the game you described (Mike being the row player and Cheng the column player), isn't it obvious that Mike is better off playing '1' when Cheng plays '0' (the lower left cell), because he receives $2$ instead of losing? $$\begin{array}{|c|c|c|}\hline&\text{'0'}&\text{'1'}\\\hline\text{'0'}&-1,1&2,-2\\\hline\text{'1'}&2,-2&-3,3\\\hline\end{array}$$ This shows precisely that it's not always optimal for Mike to play '0', since playing '1' is sometimes (when Cheng plays '0') better. $\endgroup$
    – Herr K.
    Commented Jul 6, 2017 at 18:06

1 Answer 1

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Suppose Mike plays a randomised strategy of playing "0" with probability $\dfrac58$ and playing "1" with probability $\dfrac38$

Then @Herr K.'s table becomes (in expectation) $$\begin{array}{|c|c|c|}\hline&\text{'0'}&\text{'1'}\\\hline\text{Mike's }\frac58,\frac38 \text{ combo}&\frac18,-\frac18&\frac18,-\frac18\\\hline\end{array}$$

and there is nothing Cheng can do to get a better result than an expected loss of $\frac18$. So in that sense the game is advantageous to Mike if Mike takes this approach

Incidentally, here Cheng can restrict expected losses to this amount by playing the same mixed strategy, and both players doing so is a Nash equilibrium

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  • $\begingroup$ It is not entirely incidental. The game is zero-sum and the payoff matrix is symmetric. This is enough to guarantee symmetric equilibrium. $\endgroup$
    – Giskard
    Commented Jul 15, 2017 at 5:17

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